Defining the PV circuits’ maximum and continuous current

Starting at the PV array, you can look at the individual circuits and size the conductors as they work their way to the inverter; these circuits are known collectively as the PV circuits. To size the conductors in these circuits, you have to define the maximum amount of current they’ll ever carry.

To define the maximum current that’ll ever be produced by the PV circuits (and therefore the maximum current that’ll be carried by the conductors), you need to look at Article 690.8(A)(1) of the NEC^®, titled “Circuit Sizing and Current.” The maximum current value is as follows:

The number of modules (or strings) wired in parallel × The short circuit current, I_sc, rating of the module (or string) × 1.25

Note: The 1.25 factor is in place to recognize that PV modules produce current based on the available irradiance (irradiance is the intensity of the solar radiation striking earth; flip to Chapter 4 for more information). The NEC^® also recognizes that a PV array can (and will) produce more current than its modules are rated for if given the opportunity.

If you keep reading to Article 690.8(B)(1) of the NEC^®, you’ll see that a second safety factor of 1.25 is applied. It’s there because the NEC^® dictates that the conductors and overcurrent protection devices (OCPDs) used in electrical systems “shall not continuously carry more than 80 percent of the maximum current.” The NEC^® is calling for you to oversize the conductors so that when current is run continuously (more than three hours at a time), the conductors won’t overheat, fail, and cause fires.

To make sure the conductors and OCPDs don’t carry more than the predetermined limit of 80 percent of the maximum current, you need to multiply the maximum current by 1.25 (1.25 is the inverse of 80 percent). Doing so tells you the continuous current requirement.

If you want to take a shortcut, you can multiply the two 1.25 safety factors together first to find out the overall safety factor to apply to PV circuits, which is 1.56 (1.25 × 1.25 = 1.56).

At this point, all you’ve done is define the ampacity (amount of current flow) required by the conductors on the PV side of the system. As I explain in the later “Putting together the details to determine conductor sizing” section, you have to adjust this ampacity value based on the locations of the conductors.

For example, if you have an array that consists of ten modules in series with each module short circuit current rated at 5.5 A, the maximum current from that series string of ten modules will be 5.5 A × 1.25 = 6.9 A. To define the ampacity required for the conductor connected to this string, multiply the maximum current by 1.25 again to get 6.9 A × 1.25 = 8.6 A. These results mean that the individual string is rated at 5.5 A but that the conductors and any OCPD connected to that string must be rated for at least 8.6 A.

Calculating non-PV circuits’ maximum current

Unlike the PV circuits covered in the preceding section, inverter input circuits on battery-based systems and inverter output circuits (anything from the inverter’s AC output to the load panels) aren’t subject to varying current levels based on irradiance. The maximum current levels for battery-based inverter input circuits or any inverter output circuits are actually based on the equipment to which they’re connected. (Note: The inverter input for a grid-direct inverter is on the PV side and falls into the previous section’s requirements.)

The maximum current of an inverter is defined by its manufacturer and is dictated by the maximum power the inverter can produce and the voltage it can churn out. Look for this value on an inverter’s specification (spec) sheet listed as maximum output current, as well as in the inverter’s manual.

To calculate the ampacity required by the conductors and OCPD, you only have to apply one 1.25 factor so you don’t continuously run more than 80 percent of the maximum current across these components. Here’s the formula:

Maximum output current × 1.25

For example, a 4 kW inverter operates at 240 VAC and has a maximum current output of 16.7 A. According to the preceding formula, the conductors and OCPDs connected to this inverter’s output circuit would need a minimum rating of 16.7 A × 1.25 = 20.9 A.

For battery-based inverters, the inverter input conductors run between the batteries and the inverter (or inverters if the system you’re designing calls for more than one; I outline circumstances that call for multiple inverters in Chapter 11). When applying the NEC^® factors for these circuits, you need to consider the maximum amount of current that could flow across the wires. According to Article 690.8 of the NEC^®, the maximum current for these conductors is calculated by the following formula:

Maximum continuous power output ÷ Minimum operating voltage

Performing this calculation tells you the largest amount of current that can continuously run through the conductors.

If the specified inverter has a continuous power output rating of 3 kW and it can operate all the way down at 42 VDC, the conductors between the battery bank and the inverter need to be able to handle at least 3,000 W ÷ 42 VDC = 71.4 A.

Considering conditions of use with some handy tables

At this point, you’ve calculated the maximum amount of current each conductor will need to carry based on the NEC^® requirements for each circuit (if you haven’t, refer to the earlier sections in this chapter for help). Now you need to take the conditions of use into consideration; in other words, you have to adjust the maximum current levels you found based on the locations where the conductors will be used (on the roof, in the attic, through the basement … you get the idea) as well as how many will be run together.

As conductors heat up, their ability to pass current is reduced, which is why you need to consider the conditions that’ll increase the temperature of the conductors and reduce their ampacity values. The two conditions to account for are the number of current-carrying conductors running inside a single conduit and the ambient temperature to which the conductors are exposed.

Prepare to turn to the following four tables in the NEC^® when determining the proper application of the conditions of use:

• Table 310.15(B)(2)(a), titled “Adjustment Factors for More Than Three Current-Carrying Conductors in a Raceway or Cable”
• Table 310.15(B)(2)(c), titled “Ambient Temperature Adjustment for Conduits Exposed to Sunlight On or Above Rooftops”
• Table 310.16, titled “Allowable Ampacities of Insulated Conductors Rated 0 Through 2000 Volts, 60°C Through 90°C (140°F Through 194°F), Not More Than Three Current-Carrying Conductors in Raceway, Cable, or Earth (Directly Buried), Based on Ambient Temperature of 30°C (86°F)”
• Table 310.17, titled “Allowable Ampacities of Single-Insulated Conductors Rated 0 Through 2000 Volts in Free Air, Based on Ambient Air Temperature of 30°C (86°F)”

The following sections provide basic instruction on using each of these tables.

Tables 310.16 and 310.17: Conductor ampacity values

The values listed in Tables 310.16 and 310.17 (which are represented in Figures 13-1 and 13-2; both figures show only a portion of the tables) represent the amount of current each type and size of conductor is rated for. The columns marked “Size AWG or kcmil” represent the physical size of the conductor. AWG stands for American Wire Gauge; the AWG convention is that large numbers represent small wires and small numbers represent big wires. After 0000 (pronounced four aught), the big wires are represented by their cross-sectional area instead of the AWG convention. The other columns list different types of conductors by abbreviation.

The ampacity values are based on an ambient air temperature of 30 degrees Celsius (86 degrees Fahrenheit) and no more than three current-carrying conductors (the ground wire isn’t a current-carrying conductor) run together. You adjust these values when applying the appropriate conditions of use.

Note that the tables listing the conductor ampacities also include different columns labeled with different temperatures. These are the temperature limitations of the conductors due to the insulation protecting the conductive material. Table 310.16 in Figure 13-1 differentiates between copper and aluminum conductors. Aluminum has less ampacity than copper, so to carry amounts of current that are equal to their copper counterparts, aluminum conductors must be physically larger. In all the examples I give in this book (and in most PV installations), copper conductors are used exclusively.

Conductor sizes are also limited by the rating of all the terminals, connection points where a conductor is physically connected to a termination such as a disconnect, fuse holder, or circuit breaker, that the wires are connected to. These terminals are where the conductors from the PV system connect inside the various combiner boxes, disconnects, inverters, and other pieces of electrical equipment. When you make calculations related to terminals, be sure to use Table 310.17.

The equipment used in PV systems that are less than 600 V (which are all the residential and commercial systems out there) is either 60 degrees Celsius (140 degrees Fahrenheit) or 75 degrees Celsius (167 degrees Fahrenheit). This fact means that when you use conductors rated at 90 degrees Celsius (194 degrees Fahrenheit), you need to evaluate the rating of those conductors at the terminal’s maximum value. Occasionally, this check will require you to size the conductor even larger to accommodate the temperature limitations of the terminal. The majority of the terminals used in equipment associated with PV systems are 75 degrees Celsius (167 degrees Fahrenheit), but don’t assume this is the case. Instead, verify the proper ratings so you don’t get caught installing the wrong equipment. To avoid any possible issues in the field, I suggest you only install conductors that are rated for 90 degrees Celsius (194 degrees Fahrenheit). Doing so gives you the maximum flexibility in sizing and appropriately installing the conductors.

When installing equipment with different ratings, you must use the lowest rating. For example, if the conductors are connected to a disconnect on one end with terminals rated at 60 degrees Celsius (140 degrees Fahrenheit), and a combiner box at the other end with 75 degrees Celsius (167 degrees Fahrenheit) terminals, you must use the 60 degrees Celsius terminals for the basis of your conductors’ size.

Putting together the details to determine conductor sizing

To calculate the required conductor sizes, follow these steps:

1. Determine the maximum circuit current.

   See the earlier “Defining the PV circuits’ maximum and continuous current” section for the how-to.

2. Calculate the continuous current requirement.

   For help determining the continuous current requirement, refer to the related section earlier in this chapter. (Note that the continuous current requirement defines the OCPD value you need to use in the later “Sizing Overcurrent Protection Devices and Disconnects” section.)

3. If the conductors will be installed in conduit and there will be more than three current-carrying conductors in the conduit, divide the Step 2 value by the appropriate derate value from Table 310.15(B)(2)(a); if the conductors will be exposed to sunlight on the roof, add the appropriate temperature adjustment to the ASHRAE ambient temperature value from Table 310.15(B)(2)(c), based on the height of the conductors off of the roof surface.

4. Divide the value in Step 3 by the appropriate temperature-correction factor for the expected temperature the conductors will be exposed to from the 90 degrees Celsius (194 degrees Fahrenheit) column in Table 310.16 to find the corrected conductor ampacity required.

   Choosing the temperature-correction factor from the 90 degrees Celsius (194 degrees Fahrenheit) column assumes you’re installing only conductors with 90 degrees Celsius (194 degrees Fahrenheit) ratings, as I suggest in the preceding section. If you want to use conductors rated at 75 degrees Celsius (167 degrees Fahrenheit), be sure to use that column of Table 310.16 instead.

5. Use the value from Step 3 to find the smallest 90 degrees Celsius/194 degrees Fahrenheit (or 75 degrees Celsius/167 degrees Fahrenheit) conductor that exceeds the ampacity requirement in Table 310.16.

6. Verify that the conductor selected has a large enough ampacity based on the lowest terminal ratings that will be used.

   To do this, look at the ampacity value listed in the appropriate column in Table 310.17 (which consists of terminal temperature listings) and check that the ampacity value listed for the chosen conductor is greater than or equal to the continuous current value in Step 2.

I know how tricky these calculations can be to make, so I want to share an example with you. For the purposes of this example:

• The array has eight strings in parallel, and each string has an I_sc of 5.9 A.
• All the strings are placed in parallel inside a combiner box.
• The THWN-2 conductors are in conduit and exposed to temperatures of 45 degrees Celsius, but they aren’t exposed to sunlight on a roof.
• The terminals inside the combiner box, as well as the disconnect on the other end of the conductors, are all rated at 75 degrees Celsius.

With this information, you can apply the previously described steps to determine the minimum conductor size necessary to satisfy the NEC^®:

1. Find the maximum current.

   8 strings in parallel × 5.9 A per string × 1.25 = 59 A

2. Determine the continuous current

   59 A × 1.25 = 73.75 (round up to 74 A)

3. Derate for more than three current-carrying conductors.

   In this example, all the PV source circuits are placed in parallel and a single PV output circuit leaves the combiner box. Therefore, no derate factor is applicable.

4. Apply the conditions of use.

   74 A ÷ 0.87 = 85 A

   You find the appropriate temperature derate factor from the bottom of Table 310.16. The 90 degrees Celsius (194 degrees Fahrenheit) column, which you use for THWN-2 conductors, has a temperature derate of 0.87 at 45 degrees Celsius.

   Note: If the conductors were exposed to sunlight on a roof, you’d also used a factor from Table 310.15(B)(2)(c).

5. Check Table 310.16, using the value from Step 3.

   4 AWG in the 90 degrees Celsius (194 degrees Fahrenheit) column at the top of Table 310.16 is the smallest conductor that exceeds 85 A. A THWN-2 conductor therefore has an ampacity value of 95 A.

6. Confirm that the conductor ampacity is large enough.

   According to Table 310.17, 4 AWG conductors rated at 75 degrees Celsius (167 degrees Fahrenheit) — the terminal ratings — have an ampacity of 85 A. This value exceeds the continuous current value of 74 A from Step 2. You now know that a 4 AWG conductor installed in this location meets the NEC^® requirements for ampacity and can safely carry current from this array.

You can follow this process for all the circuits used within a PV system. When calculating for circuits other than those on the PV side (the inverter output circuit, for example), you start at Step 1 with the maximum current value as given by the manufacturer as I describe in the earlier “Calculating non-PV circuits’ maximum current” section. You then proceed through the rest of the steps as normal.

Accounting for voltage drop after you size your conductors

The NEC^® requires that the conductors you use in any PV system must be large enough to carry current from the source to the load safely, but no NEC^® requirement dictates that you must deliver the power as efficiently as possible. As a PV system designer and installer, though, you want to make sure that the power supplied by the PV system is never reduced because the conductors are too small.

No matter how large you size the conductors in your system (see the previous section for guidelines), some amount of voltage drop will occur. As current flows in the conductors, resistance in the conductors fights that current flow. As you discover in Chapter 3, voltage has a direct relationship to current (which is measured in amps) and resistance via Ohm’s Law, which can be stated in several ways:

• Volts = Amps × Resistance (E = I × R)
• Amps = Volts ÷ Resistance (I = E ÷ R)
• Resistance = Volts ÷ Amps (R = E ÷ I)

This interplay between current flow and resistance means you need to evaluate the conductors used in your system and make sure they aren’t so small that excessive resistance causes too much loss.

Because the NEC^® doesn’t dictate the maximum amount of voltage drop, you get to decide which amount suits you best. Within the solar industry, the level of voltage drop is referred to in terms of the DC and AC sides of the system.

• The accepted amount of DC voltage drop is 2 percent from the PV array down to the inverter input. This isn’t a hard value, though. If you want to, you can evaluate the financial effects of running larger conductors to reduce voltage drop and find that a 3-percent drop makes sense.
• A tolerable amount of voltage drop on the AC side is between 1 percent and 1.5 percent for utility-interactive systems (either grid-direct or battery-based). If the PV system is of the stand-alone, battery-based variety, between 2 percent and 3 percent voltage drop is okay.

I show you how to calculate the voltage drop values for DC and AC circuits in the next sections. In both examples, I use Ohm’s Law and some given properties for copper conductors (but you can solve for any of the pieces of information as long as you can define all the other parts of the equation). Note: In the real world, you may need to take slight differences in the nature of AC circuits into account when calculating the AC voltage drop. For all practical purposes, though, using Ohm’s Law as I show you is accurate enough.

If you calculate the voltage drop and come up with a gauge that’s different than the calculations in the preceding section, use the larger of the two conductors. This way you always satisfy both requirements.

DC voltage

The first step in calculating the voltage drop from the PV array down to the inverter is to break down the individual parts of the PV circuit(s).

To estimate the total voltage drop, you need to consider the longest PV source circuit length (from one of the strings to the combiner) and the PV output circuit (from the combiner to the inverter) and make two separate calculations that you then add together.

Note: By considering the longest PV source circuit length, you’ll be slightly more conservative for the shorter lengths, which results in slightly better performance in those circuits.

Before you can run your calculations, you need to collect the following information about the PV system:

• The maximum power output rating of the array from the modules’ spec sheets: You use the V_mp value because you’re most concerned about the voltage drop when the array is producing its greatest amount of power.
• The maximum current value for each circuit from the modules’ spec sheets: You use the I_mp value for the same reason you use the V_mp value.
• The total circuit length: This number tells you how far the electrons have to travel to complete their circuit. You generally have to estimate this number by measuring the distances for the proposed equipment locations. Convert this distance into kilofeet (a goofy measurement, I know) because the table you have to reference supplies conductor resistances in terms of resistance per thousand feet of conductor. To convert to kilofeet, divide the estimated distance by 1,000.
• The resistance of the conductor: Refer to the “Conductor Properties” table in the NEC^® (a portion of which is in Figure 13-3) to see the different resistance characteristics of conductors. When you have an idea of the conductor size to use (because you did the NEC^® calculations in the previous section or because you’re installing modules with preinstalled conductors), go to this table and use the resistance per kilofeet to swap into the equation and solve for voltage drop.

Calculating the DC voltage drop requires you to apply the information you collected regarding the module specs and the conductor properties (length and resistance) to a form of Ohm’s Law a little differently than Herr Ohm intended (although if given the chance, I think he’d approve). The difference is that a length value is associated with the equation:

• V_drop = I_mp × R_c × L
• V_drop = The maximum number of volts you’re willing to lose based on the array’s V_mp (it’s the variable you solve for)
• I_mp = The maximum power current rating of the circuit
• R_c = The resistance of the conductor (Ω/kft)
• L = The total circuit length in thousands of feet (kft)

Suppose you have an array that consists of three strings of ten modules and all the strings are arranged on a roof in rows of ten. These three strings are all wired to a roof-mounted combiner box that runs down from the roof to the inverter mounted outside next to the utility meter and MDP. This setup means you have three PV source circuits running from the strings to the combiner box and a single circuit running the rest of the way out.

Each PV module is 36.2 V and 4.9 A, and the strings are wired with 12 AWG conductors. With all of this information, you can measure the total circuit length for the circuit and calculate the total voltage drop. In order to determine the voltage drop in the PV source circuit, you need to collect the following module and conductor information:

• V_mp for the modules is 36.2 V, so the string V_mp = 362 V.
• I_mp for the module is 4.9 A, which means the string I_mp is also 4.9 A.
• The total circuit length is 75 feet, or 75 ÷ 1,000 = 0.075 kilofeet.
• The resistance, R_c, in the 12 AWG wire used is 1.98 Ω/kilofeet (from the NEC^® table).

Here’s how to calculate the number of volts that will be lost in this circuit (in other words, the voltage drop, or V_drop), using the preceding information:

• V_drop = I_mp × R_c × L
• V_drop = 4.9 A × 1.98 Ω/kilofeet × 0.075 kilofeet
• V_drop = 0.73 V

This number indicates how many volts will be lost due to the resistance in the conductor. Your next step is to relate this voltage drop to a percentage of the entire circuit. To determine the voltage drop in a percentage, divide the calculated loss by the original V_mp value for the array:

V_drop % = V_drop / V_mp of the array

So when you apply the values calculated in the example, you get the following:

0.73 V ÷ (36.2 V × 10) = 0.73 V ÷ 362 V = 0.002 = 0.2%

This result indicates that a 0.2-percent voltage drop exists in the PV source circuits. To determine the voltage drop in the PV output circuit, repeat this process. If when you add these two voltage drop percentages together, the total voltage drop is less than your desired amount of 2 percent, the conductor sizes check out. If the total voltage drop is greater than your desired amount, you need to upsize one of the conductors to minimize the loss.

Beginning with a few basics

Believe it or not, in the process of sizing the PV system’s conductors, you also defined the required size of the OCPDs for the conductors. (It’s based off the continuous current requirement in Step 2 of the process described in the earlier “Putting together the details to determine conductor sizing” section.)

If you can find an OCPD that’s listed for 100 percent of its rating, use the maximum current rating of your circuit as the size of your OCPD (Step 1 of the conductor sizing process). OCPDs with 100 percent ratings aren’t common, though. Check the spec sheet provided by the manufacturer to verify whether a particular OCPD can really handle as much current as an array produces.

When you calculate an OCPD size, most likely it’ll be a value that isn’t commonly available. Fortunately, the NEC^® allows you to round up to the next available OCPD size. These commonly available sizes can be found in Article 240.6 of the NEC^®. In the example I use earlier to show you how to calculate the conductor size, the continuous current value was 74 A. OCPDs sized at 74 A aren’t regularly available, so the OCPD size you’d need to use for the circuit in question would be rated at 80 A (the next available size).

If the circuit uses small conductors, be aware that the NEC^® limits the size of OCPD you can connect these wires to, regardless of the conductors’ ampacity. As reported in Article 240.4(D) of the NEC^®, the limitations are as follows:

• 14 AWG connected to 15 A or less
• 12 AWG connected to 20 A or less
• 10 AWG connected to 30 A or less

To put these limitations in perspective, if you run the calculations and determine that a 35 A OCPD is required for your installation, then you can’t use a 10 AWG conductor, even though in some situations that conductor has an ampacity value of 40 A.

Manufacturers of PV modules and inverters are required to specify the maximum OCPD their equipment can be connected to, known as the maximum series fuse rating for PV modules. It’s listed with every module you ever buy. So when you’re buying OCPDs for the PV strings, know that the manufacturer has defined the largest amperage rating its OCPD can have.

Placing protection on PV circuits

Although you may be used to placing OCPD on each and every circuit you wire in a regular electrical system, placing OCPD in every PV source and output circuit isn’t an absolute requirement. When your system meets Article 690.9 of the NEC^®, you can actually place two strings of PV modules in parallel without having to put OCPDs on either string. As soon as you add a third string, though, you should place OCPDs on the strings.

To meet NEC^® requirements eliminating OCPDs on PV circuits, you first have to size your conductors to meet Article 690.8. Next, you can’t have any potential backed currents, current that can come back from sources such as batteries and inverters. (This is true when using grid-direct inverters, but if you’re using battery-based inverters, you won’t meet the requirements, so you have to use OCPD on all PV circuits.) Finally, the short circuit current (I_sc) from all sources can’t exceed the ampacity of the conductors. (This last requirement is where the limit of two strings in parallel comes into play.)

The idea is, if you have three strings in parallel and a fault occurs on one of the strings (meaning that string becomes a load and starts accepting current), the I_sc values from the other two strings (the external sources) will exceed the conductors’ ampacity and the modules’ rating.

Follow this general rule for grid-direct systems: When you have one or two strings, skip the OCPD; when you have more than two strings, put an OCPD on each string. (Note that battery-based systems require an OCPD on every string.)

Rooftop combiner boxes are a common and acceptable location for series string OCPDs. These devices are affected by temperature though, so check the manufacturer’s specifications for proper OCPD sizing when installing an OCPD in high-temperature locations (like rooftops).

Protecting inverter circuits

As for the AC output of your inverter, you always need an OCPD for this circuit. A variety of options exist for installing this protection, with the most common being a circuit breaker located in the MDP. This circuit breaker serves as both a way to disconnect the inverter and a method of protecting the conductors. (Some installations call for a fused AC disconnect to be installed rather than a circuit breaker; I cover such installations in Chapter 17.)

To determine the amperage rating of the OCPD connected to the inverter, you need to refer to the inverter manufacturer’s specifications. Many inverter manufacturers list this value on their spec sheets. If you encounter one that doesn’t, you may need to dig into the installation manual; if nothing else, that should indicate the maximum OCPD sizing you can use in conjunction with the inverter.